Replacing an operator of an angular momentum for a constant.

In summary, the conversation discusses replacing an operator of the ##z## component of angular momentum with its expectation value in a spherical symmetric potential. The equations presented show that the function ##\psi## is analytic in ##\varphi## and behaves as ##e^{im\varphi}##, which are eigenfunctions of the angular momentum operator.
  • #1
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While dealing with a circling particle in an spherical symetric potential our professor said that we can replace an operator of ##z## component of angular momentum ##\hat{L}_z## with the expectation value - he denoted it just ##L_z## - of the angular momentum if ##L_z## is constant. Why is that?

So we first had this equation:

\begin{align}
\underbrace{\psi (r,\varphi,\vartheta)}_{\rlap{\text{w. f. in spherical coordinates}}} &= \exp\left[\hat{L}_z \frac{i}{\hbar}\, \varphi\right] \underbrace{\psi (r,0,\vartheta)}_{\rlap{\text{w. f. in spherical. coordinates at $\varphi=0$}}}
\end{align}

and we got this one (notice that there is no operator over an ##L_z##):

\begin{align}
\psi (r,\varphi,\vartheta) &= \exp\left[L_z \frac{i}{\hbar}\, \varphi\right] \psi (r,0,\vartheta)
\end{align}

Anyway here is the spherical coordinate system we ve been using all the time (the blue spherical aure is supposed to be a spherical potential...):

G8aUE.png
 
Last edited:
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  • #2
I think there should be imaginary unit inside the[ ] brackets. The first equation states that the function ##\psi## is analytic in ##\varphi##. The second equation states more than that: it says that the function behaves as ##e^{im\varphi}##, i.e. oscillates. Such functions are eigenfunctions of the angular momentum operator, if the number ##m = l_z/\hbar## is whole number.
 
  • #3
You are soo right! I misstyped ##i## for ##1##. I corrected that mistake. I don't think i understand your explanation though.
 

1. What is an operator of angular momentum?

An operator of angular momentum is a mathematical representation of the physical quantity of angular momentum in quantum mechanics. It is used to describe the rotational motion of particles and is an important concept in understanding the behavior of atoms and molecules.

2. Why would an operator of angular momentum need to be replaced with a constant?

In certain situations, it may be necessary to replace an operator of angular momentum with a constant value. This is often done when studying simpler systems that do not exhibit the complex rotational behavior that requires the use of the operator.

3. What is the process for replacing an operator of angular momentum with a constant?

The process for replacing an operator of angular momentum with a constant involves simplifying the mathematical expression for the operator by removing terms that are not necessary for the specific system being studied. This results in a constant value that represents the total angular momentum of the system.

4. What are the implications of replacing an operator of angular momentum with a constant?

Replacing an operator of angular momentum with a constant can make the mathematical analysis of a system easier and more manageable. It allows for a better understanding of the behavior of the system and can lead to more accurate predictions of its properties.

5. Are there any limitations to replacing an operator of angular momentum with a constant?

While replacing an operator of angular momentum with a constant can be useful in certain situations, it is important to note that this simplification may result in a loss of information about the system. It is not always possible to accurately represent the complex rotational behavior of a system with a single constant value.

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